Current Divider: Formula, Rule, Conductance & Examples

A current divider circuit consists of two or more parallel branches that provide multiple paths for electric current to flow. In a parallel circuit, the voltage across each branch is the same, while the total current divides among the branches according to their resistance. A branch with lower resistance carries more current, whereas a branch with higher resistance carries less current.

The Current Divider Rule

A current divider is a parallel circuit where the supply current splits between two or more parallel branches. Each branch provides a separate path for current, while all branches are connected across the same two nodes.

In a parallel circuit, the voltage is the same across every branch, but the current through each component can be different depending on its resistance. Therefore:

VR1=VR2=VR3=V_{R1} = V_{R2} = V_{R3} = \cdots

The individual branch currents can be determined easily using Ohm’s Law and Kirchhoff’s Current Law (KCL). This principle forms the basis of the current divider rule used for analyzing parallel resistor circuits.

Understanding Current Division

The simplest current divider circuit consists of two resistors connected in parallel. In this arrangement, the total current divides between the two branches according to their resistance.

The Current Divider Rule provides a quick way to calculate the current flowing through each parallel resistor as a portion of the total circuit current. It is especially useful for analyzing parallel circuits without first calculating the voltage across each resistor.

Consider the parallel resistive network below to understand how current division works.

Current Dividing Parallel Circuit

Current Dividing Parallel Circuit

A basic current divider circuit consists of two resistors, R1R_1 and R2R_2, connected in parallel. The source current ITI_T divides at the junction into two branch currents, IR1I_{R1} and IR2I_{R2}, which combine again before returning to the source.

According to Kirchhoff’s Current Law (KCL), the total current entering the parallel network is equal to the sum of the branch currents:

IT=IR1+IR2I_T = I_{R1} + I_{R2}

Therefore, the current in either branch can be found from the total current and the current in the other branch:

IR1=ITIR2I_{R1} = I_T – I_{R2}
IR2=ITIR1I_{R2} = I_T – I_{R1}

Because R1R_1 and R2R_2 are connected in parallel, the same voltage VV appears across both resistors. Using Ohm’s Law, the branch currents can therefore be expressed as:

IR1=VR1I_{R1} = \frac{V}{R_1}
IR2=VR2I_{R2} = \frac{V}{R_2}

This common-voltage relationship forms the basis for deriving the current divider formula. Therefore, using the equivalent resistance of the two parallel resistors, the voltage across the parallel combination can be expressed as:

 voltage (V) across the parallel combination of resistors in current divider circuit

Solving for IR1I_{R1}  gives:

Soving for IR1 in current division- current division formula

Similarly, solving for IR2I_{R2} gives:

solving for IR2 in current divider

Notice that each branch-current equation contains the opposite resistor in the numerator. Thus, I1I_1 uses R2R_2, while I2I_2 uses R1R_1. This is because branch current is inversely proportional to resistance, so the lower-resistance branch carries the greater current.

Current Divider Worked Example No. 1

A 15 Ω resistor is connected in parallel with a 45 Ω resistor. The parallel combination is connected across a 24 V battery supply. Calculate:

  1. The current flowing through each resistor.
  2. The total current supplied by the source.

Using the values R1R_1, R2R_2, and V=24VV=24 V:

IR1=VR1=2415=1.6AI_{R1} = \frac{V}{R_1} = \frac{24}{15} = 1.6\,\text{A}
IR2=VR2=2445=0.533AI_{R2} = \frac{V}{R_2} = \frac{24}{45} = 0.533\,\text{A}
IT=IR1+IR2=1.6+0.533=2.133AI_T = I_{R1} + I_{R2} = 1.6 + 0.533 = 2.133\,\text{A}

The 15 Ω resistor carries more current than the 45 Ω resistor because current is inversely proportional to resistance. In a parallel circuit, the lower-resistance branch always carries the greater current. A short circuit with nearly zero resistance can therefore allow very high current, while an open circuit with extremely high resistance carries essentially no current.

The equivalent resistance REQR_{\text{EQ}} of parallel-connected resistors is always less than the smallest individual resistance. Adding more parallel branches further reduces the equivalent resistance.

If the total current ITI_T is already known, it is not always necessary to calculate every branch current. The remaining branch current can be found by subtracting the known branch currents from the total current, according to Kirchhoff’s Current Law (KCL).

Current Divider Worked Example No. 2

Three resistors are connected together to form a current divider circuit as shown below. If the circuit is supplied from a 120 V source with a power capacity of 1.8 kW, calculate the individual branch currents using the current divider rule and determine the equivalent circuit resistance.

Current Divider Worked Example No2

Finding Equivalent Resistance

The equivalent resistance of the three parallel resistors is first calculated as:

Req=1120+140+160R_{\text{eq}} = \frac{1}{\frac{1}{20} + \frac{1}{40} + \frac{1}{60}}
Req=10.91ΩR_{\text{eq}} = 10.91\,\Omega

Therefore, the total circuit current is:

IT=VSReqI_T = \frac{V_S}{R_{\text{eq}}}
IT=12010.91=11AI_T = \frac{120}{10.91} = 11\,\text{A}

Branch currents IR1, IR2, IR3I_{R1},\ I_{R2},\ I_{R3}

Using the current divider rule:

IR1=ITReqR1I_{R1} = I_T \frac{R_{\text{eq}}}{R_1}
IR1=11×10.9120=6AI_{R1} = 11 \times \frac{10.91}{20} = 6\,\text{A}

Similarly,

IR2=ITReqR2=11×10.9140=3AI_{R2} = I_T \frac{R_{\text{eq}}}{R_2} = 11 \times \frac{10.91}{40} = 3\,\text{A}
IR3=ITReqR3=11×10.9160=2AI_{R3} = I_T \frac{R_{\text{eq}}}{R_3} = 11 \times \frac{10.91}{60} = 2\,\text{A}

Therefore:

IT=IR1+IR2+IR3I_T = I_{R1} + I_{R2} + I_{R3}
IT=6+3+2=11AI_T = 6 + 3 + 2 = 11\,\text{A}

We can verify the results using Kirchhoff’s Current Law (KCL). The total current must equal the sum of the three branch currents:

IT=IR1+IR2+IR3=6+3+2=11AI_T = I_{R1} + I_{R2} + I_{R3} = 6 + 3 + 2 = 11\,\text{A}

This confirms our calculation. The total current is divided among the parallel branches according to their resistance values.

For a given supply voltage, adding more resistors in parallel generally increases the total supply current, because the equivalent resistance of the circuit decreases as additional current paths are added.

Current Division Using Conductances

Another useful method for finding branch currents in a DC parallel circuit is the conductance method. In a parallel network, conductance indicates how easily each branch allows electric current to flow. It is represented by the letter G and is the reciprocal of resistance.

Resistance is measured in ohms (Ω), while conductance is measured in siemens (S). The older term mho (℧) is also used for conductance; the symbol is simply an inverted ohm symbol. The siemens is the standard unit used in modern electrical engineering.

For resistors connected in parallel, the total or equivalent conductance is equal to the sum of the conductances of the individual branches. This makes the conductance method particularly convenient for analyzing current distribution in parallel circuits.

Parallel Conductance of a Current Divider

The formula of Parallel Conductance of a Current Divider is:

1RT=1R1+1R2+1R3+\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots
GT=G1+G2+G3+G_T = G_1 + G_2 + G_3 + \cdots

Conductance provides a convenient way to analyze how current is distributed among parallel branches. For example, a resistor with a resistance of 20 Ω has a conductance of 0.05 S. Because conductance and resistance are reciprocals, a higher conductance corresponds to lower resistance, while a lower conductance corresponds to higher resistance.

Conductance can also be expressed using SI prefixes such as millisiemens (mS), microsiemens (μS), and nanosiemens (nS). For example, a 20 kΩ resistor has a conductance of 50 μS.

This relationship also allows the current divider rule to be expressed using conductance instead of resistance, providing another convenient method for calculating individual branch currents.

Using Ohm’s Law, current is expressed as voltage divided by resistance. Since conductance is the reciprocal of resistance, the branch current can instead be expressed as the product of voltage and conductance:

I=V×GI = V \times G

For a parallel resistive network, the same voltage is present across every branch. Therefore, the total supply current depends on the combined conductance of all the parallel branches.

Because voltage can also be expressed in terms of current and conductance, we can write:

V=IGV = \frac{I}{G}

Using these relationships, the current divider rule can be expressed in terms of conductance GG rather than resistance RR.

Current Divider Rule using Conductance

IR1=G1×V=G1(ITGT)I_{R1} = G_1 \times V = G_1\left(\frac{I_T}{G_T}\right)
IR1=IT(G1GT)\therefore I_{R1} = I_T\left(\frac{G_1}{G_T}\right)

Similarly, the currents through the parallel resistors R2R_2 and R3R_3 can be expressed as:

IR2=IT(G2GT);IR3=IT(G3GT)I_{R2} = I_T\left(\frac{G_2}{G_T}\right); \qquad I_{R3} = I_T\left(\frac{G_3}{G_T}\right)

Unlike the resistance-based current divider equations, the same branch conductance appears in the numerator of each conductance-based equation. Thus, I1I_1 is calculated using G1G_1 , while I2I_2 is calculated using G2G_2 . This is because conductance is the reciprocal of resistance, so a higher conductance corresponds to a lower resistance and a greater share of the total current.

Current Dividers Worked Example No3

Using the conductance method, calculate the individual branch currents I1I_1, I2I_2, and I3I_3 in the following parallel resistive circuit.

Current Dividers Worked Example No3 circuit diagram

Total conductance GTG_T

GT=1R1+1R2+1R3G_T = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
GT=13000+16000+112000G_T = \frac{1}{3000} + \frac{1}{6000} + \frac{1}{12000}
GT=4+2+112000G_T = \frac{4 + 2 + 1}{12000}
GT=712000G_T = \frac{7}{12000}
GT=0.0005833S=583.3μS\therefore G_T = 0.0005833\,\text{S} = 583.3\,\mu\text{S}

Total supply current ITI_T

IT=VS×GTI_T = V_S \times G_T
IT=24×0.0005833=0.014AI_T = 24 \times 0.0005833 = 0.014\,\text{A}
IT14mAI_T \approx 14\,\text{mA}
G1=13000=333.3μSG_1 = \frac{1}{3000} = 333.3\,\mu\text{S}
G1=13000=333.3μSG_1 = \frac{1}{3000} = 333.3\,\mu\text{S}
G2=16000=166.7μSG_2 = \frac{1}{6000} = 166.7\,\mu\text{S}
G3=112000=83.3μSG_3 = \frac{1}{12000} = 83.3\,\mu\text{S}

Individual branch currents I1, I2, and I3I_1,\ I_2,\ \text{and}\ I_3

IR1=IT(G1GT)I_{R1} = I_T\left(\frac{G_1}{G_T}\right)
IR1=0.014(0.00033330.0005833)=0.008A=8mAI_{R1} = 0.014\left(\frac{0.0003333}{0.0005833}\right) = 0.008\,\text{A} = 8\,\text{mA}
IR2=IT(G2GT)I_{R2} = I_T\left(\frac{G_2}{G_T}\right)
IR2=0.014(0.00016670.0005833)=0.004A=4mAI_{R2} = 0.014\left(\frac{0.0001667}{0.0005833}\right) = 0.004\,\text{A} = 4\,\text{mA}
IR3=IT(G3GT)I_{R3} = I_T\left(\frac{G_3}{G_T}\right)
IR3=0.014(0.00008330.0005833)=0.002A=2mAI_{R3} = 0.014\left(\frac{0.0000833}{0.0005833}\right) = 0.002\,\text{A} = 2\,\text{mA}

Since conductance is the reciprocal of resistance, the equivalent resistance of the example circuit can be obtained by taking the reciprocal of the total conductance:

Req=1GTR_{\text{eq}} = \frac{1}{G_T}
Req=1583.3μSR_{\text{eq}} = \frac{1}{583.3\,\mu\text{S}}
Req1714ΩR_{\text{eq}} \approx 1714\,\Omega
Req1.71kΩR_{\text{eq}} \approx 1.71\,\text{k}\Omega

This value is lower than the smallest resistor, R1=3kΩR_1 = 3\,\text{k}\Omega, as expected for resistors connected in parallel.

Conclusion

A current divider is a parallel circuit in which the total current divides among the individual branches, while the same voltage is present across each parallel element. Kirchhoff’s Current Law (KCL) states that the total current entering a junction is equal to the sum of the currents leaving it.

When two parallel resistors have the same resistance, the total current divides equally between them. When their resistance values are different, the branch with lower resistance carries more current, while the branch with higher resistance carries less.

For circuits with three or more parallel branches, the equivalent resistance can be used along with the total current to determine the current in each branch. The current distribution depends on the inverse of the branch resistance, and the total current is the sum of all branch currents.

The conductance method provides another convenient approach for current division. Since conductance is the reciprocal of resistance, the total conductance of a parallel network is the sum of the individual branch conductances. Conductance is measured in siemens (S) and can be used to analyze current division in both DC and AC circuits.

Read Next:

  1. Resistors in Parallel – Parallel Connected Resistors Explained
  2. Resistors in Series: Definition, Formula, Rules, Calculation, and Examples
  3. Capacitors in Series and Series Capacitor Circuits – Complete Guide
  4. Voltage Divider: Rule, Equations, Formulas, and Practical Applications
  5. AC Capacitance: Formula, Reactance & Phase Angle
  6. Non-inverting Operational Amplifier: Circuit, Gain & Formula

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